\(\dfrac{3}{5x-5}-\dfrac{x+5}{10x-10}\)
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ND
11 tháng 12 2017
\(a.\)
\(\left(x^2-25\right):\dfrac{2x+10}{3x-7}\)
\(=\left(x-5\right)\left(x+5\right).\dfrac{3x-7}{2\left(x+5\right)}\)
\(=\dfrac{\left(x-5\right)\left(x+5\right)\left(3x-7\right)}{2\left(x+5\right)}\)
\(=\dfrac{\left(x-5\right)\left(3x-7\right)}{2}\)
\(b.\)
\(\dfrac{x^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}\)
\(=\dfrac{x\left(x+1\right)}{5\left(x^2-2x+1\right)}.\dfrac{5\left(x-1\right)}{3\left(x+3\right)}\)
\(=\dfrac{x\left(x+1\right)}{5\left(x-1\right)^2}.\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right).5\left(x-1\right)}{5\left(x-1\right)^2.3\left(x+1\right)}\)
\(=\dfrac{x}{3\left(x-1\right)}\)
NH
11 tháng 12 2017
\(\dfrac{x^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}=\dfrac{5x\left(x+1\right)\left(x-1\right)}{15\left(x-1\right)^2\left(x+1\right)}=\dfrac{x}{3\left(x-1\right)}\)\(\left(x^2-25\right):\dfrac{2x+10}{3x-7}=\dfrac{\left(x-5\right)\left(x+5\right)\left(3x-7\right)}{2\left(x+5\right)}=\dfrac{\left(x-5\right)\left(3x-7\right)}{2}\)
14 tháng 12 2021
\(=\dfrac{x\left(x+1\right)}{5\left(x+1\right)^2}\cdot\dfrac{5x-1}{3\left(x+1\right)}=\dfrac{x\left(5x-1\right)}{15\left(x+1\right)^2}\)
14 tháng 12 2021
\(\dfrac{x^2+x}{5x^2+10x+5}:\dfrac{3x+3}{5x-1}=\dfrac{x\left(x+1\right)}{5\left(x^2+2x+1\right)}:\dfrac{3\left(x+1\right)}{5x-1}=\dfrac{x\left(x+1\right)}{5\left(x+1\right)^2}.\dfrac{5x-1}{3\left(x+1\right)}=\dfrac{x\left(5x-1\right)}{15\left(x+1\right)^2}\)
P
1
12 tháng 2 2022
a: =>10x-14=15-9x
=>19x=29
hay x=29/19
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>30x+9=32x+60
=>-2x=51
hay x=-51/2
c: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
=>35x-5+60x=96-6x
=>101x=101
hay x=1
d: \(\Leftrightarrow12\left(\dfrac{1}{2}-\dfrac{3}{2}x\right)=-5x+6\)
\(\Leftrightarrow6-18x+5x-6=0\)
=>-13x=0
hay x=0
12 tháng 2 2022
\(a,\dfrac{5x-7}{3}=\dfrac{5-3x}{2}\\ \Leftrightarrow2\left(5x-7\right)=3\left(5-3x\right)\\ \Leftrightarrow10x-14=15-9x\\ \Leftrightarrow10x-14-15+9x=0\\ \Leftrightarrow19x-19=0\\ \Leftrightarrow x=1\)
\(b,\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\\ \Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(6+8x\right)}{36}\\ \Leftrightarrow30x+9=36+24+32x\\ \Leftrightarrow36+24+32x-30x-9=0\\ \Leftrightarrow2x+51=0\\ \Leftrightarrow x=-\dfrac{51}{2}\)
\(c,\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\\ \Leftrightarrow\dfrac{7x-1+12x}{6}=\dfrac{16-x}{5}\\ \Leftrightarrow5\left(19x-1\right)=6\left(16-x\right)\\ \Leftrightarrow95x-5=96-6x\\ \Leftrightarrow95x-5-96+6x=0\\ \Leftrightarrow101x-101=0\\ \Leftrightarrow x=1\)
\(d,4\left(0,5-1,5x\right)=-\dfrac{5x-6}{3}\\ \Leftrightarrow12\left(0,5-1,5x\right)=6-5x\\ \Leftrightarrow6-18x=6-5x\\ \Leftrightarrow6-5x-6+18x=0\\ \Leftrightarrow13x=0\\ \Leftrightarrow x=0\)
DL
1
27 tháng 7 2023
Đặt x/3=y/5=k
=>x=3k; y=5k
\(A=\dfrac{5\cdot9k^2+3\cdot25k^2}{10\cdot9k^2-3\cdot25k^2}=\dfrac{5\cdot9+3\cdot25}{10\cdot9-3\cdot25}=8\)
21 tháng 1 2021
2) Ta có: \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30\left(x-4\right)}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)
\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)
\(\Leftrightarrow-24x+144=-5x+30\)
\(\Leftrightarrow-24x+144+5x-30=0\)
\(\Leftrightarrow-19x+114=0\)
\(\Leftrightarrow-19x=-114\)
hay x=6
Vậy: x=6
3) Ta có: \(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(\Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(6+8x\right)}{36}\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow30x+9-60-32x=0\)
\(\Leftrightarrow-2x-51=0\)
\(\Leftrightarrow-2x=51\)
hay \(x=-\dfrac{51}{2}\)
Vậy: \(x=-\dfrac{51}{2}\)
4) Ta có: \(\dfrac{x+1}{3}-\dfrac{x-2}{6}=\dfrac{2x-1}{2}\)
\(\Leftrightarrow\dfrac{2\left(x+1\right)}{6}-\dfrac{x-2}{6}=\dfrac{3\left(2x-1\right)}{6}\)
\(\Leftrightarrow2x+2-x+2=6x-3\)
\(\Leftrightarrow x+4-6x+3=0\)
\(\Leftrightarrow-5x+7=0\)
\(\Leftrightarrow-5x=-7\)
hay \(x=\dfrac{7}{5}\)
Vậy: \(x=\dfrac{7}{5}\)
KV
21 tháng 1 2021
1) \(\dfrac{5x-2}{3}=\dfrac{5-3x}{2}\)
\(2\left(5x-2\right)=3\left(5-3x\right)\)
\(10x-4=15-9x\)
\(10x+9x=15+4\)
\(19x=19\)
\(x=1\)
Vậy \(x=1\)
7 tháng 10 2021
a: \(5x^2y^4:10x^2y=\dfrac{1}{2}y^3\)
c: \(\left(-xy\right)^{10}:\left(-xy\right)^5=-x^5y^5\)
HC
2
19 tháng 7 2021
\(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\)
\(\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}=\dfrac{x+5}{2\left(x-5\right)\left(x+5\right)}\)
dkxd : x ≠ 0
x ≠ 5
x ≠ -5
MTC : 2x(x - 5)(x + 5)
Quy đồng mẫu thức hai vế của phương trình :
⇒ \(\dfrac{2\left(x-5\right)\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}\) = \(\dfrac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}\)
Suy ra : 2(x - 5)(x + 5) - (x - 5)(x + 5) = x(x + 25)
\(\Leftrightarrow\) 2(x2 - 25) - (x2 - 25) = x2 + 25x
\(\Leftrightarrow\) 2x2 - 50 - x2 + 25 - x2 - 25x = 0
\(\Leftrightarrow\) -25 - 25x = 0
\(\Leftrightarrow\) -25x = 25
\(\Leftrightarrow\) x = \(\dfrac{25}{-25}=-1\) (thỏa mãn)
Vậy S = \(\left\{-1\right\}\)
Chúc bạn học tốt
19 tháng 7 2021
Ta có: \(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\)
\(\Leftrightarrow\dfrac{2\left(x+5\right)^2}{2x\left(x+5\right)\left(x-5\right)}-\dfrac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}=\dfrac{x\left(x+25\right)}{2x\left(x+5\right)\left(x-5\right)}\)
Suy ra: \(2\left(x^2+10x+25\right)-\left(x^2-10x+25\right)=x^2+25x\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\)
\(\Leftrightarrow15x+25=0\)
\(\Leftrightarrow15x=-25\)
hay \(x=-\dfrac{5}{3}\)(thỏa ĐK)
\(\dfrac{3}{5x-5}-\dfrac{x+5}{10x-10}=\dfrac{6}{10\left(x-1\right)}-\dfrac{x+5}{10\left(x-1\right)}=\dfrac{6-\left(x+5\right)}{10\left(x-1\right)}\)
\(=\dfrac{-\left(x-1\right)}{10\left(x-1\right)}=-\dfrac{1}{10}\)
\(\dfrac{3}{5x-5}-\dfrac{x+5}{10x-10}\)
= \(\dfrac{3}{5.(x-1)}\) - \(\dfrac{x+5}{2.5.(x-1)}\)
= \(\dfrac{6-x-5}{10.(x-1)}\)
= \(\dfrac{1-x}{10.(x-1)}\)
= \(\dfrac{-1(x-1)}{10(x-1)}\)
= \(\dfrac{-1}{10}\)